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3 years ago

In an overhead camshaft engine, the lobes make direct contact with the

Physics

2 answers:

luda_lava [24]3 years ago

8 0

The answer would have to be a,value springs

jekas [21]3 years ago

8 0

Answer: B. Rocker arms

Explanation: In an overhead camshaft (OHC) engine,

the camshaft, rocker arms, valve springs, and valves are all

located in the cylinder head. The camshaft is driven off the

crankshaft by using a timing chain. The lobes press directly

up onto rocker arms, and the rocker arms push down on the

valves to open them

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Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstro

Oduvanchick [21]

Answer:

Explanation:

The formula for hydrogen atomic spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

$E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV$

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc

energy of first line

$E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})$

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

$E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})$

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

$E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})$

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

$E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})$

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

$E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})$

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

7 0

2 years ago

anastassius [24]

Answer:

the answer is c

Explanation:

7 0

2 years ago

Forces that are equal in size but opposite in direction are called what ?

enot [183]

Answer:

I think they are called balanced forces

Explanation:

3 0

2 years ago

Read 2 more answers

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o

MrRa [10]

Answer:

18.6012339739 A

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

$B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A$

The current necessary to produce this field is 18.6012339739 A

7 0

2 years ago

2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across

gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

4 0

1 year ago